TUTORIAL 7 MTH3201 LINEAR ALGEBRA

𝐴𝑇 =

3/7 −6/7 2/72/7 3/7 6/76/7 2/7 −3/7

𝐴𝐴𝑇 =

3/7 2/7 6/7−6/7 3/7 2/72/7 6/7 −3/7

3/7 −6/7 2/72/7 3/7 6/76/7 2/7 −3/7

=1 0 00 1 00 0 1

= 𝐼

𝐴𝐴𝑇 = 𝐼 𝐴𝐴𝑇 = 𝐴𝐴−1

∴ 𝐴𝑇= 𝐴−1

Since , 𝐴 is an orthogonal matrix 𝐴𝑇 = 𝐴−1

1(a)

𝑅𝑜𝑤 𝑠𝑝𝑎𝑐𝑒 𝑜𝑓 𝐴: 𝑟 1 = 3/7 2/7 6/7 𝑟 2 = −6/7 3/7 2/7

𝑟 3 = 2/7 6/7 −3/7

𝑟 1, 𝑟 2 = 3/7 −6/7 + (2/7)(3/7) + 6/7 2/7 = 0

𝑟 1, 𝑟 3 = 3/7 2/7 + (2/7)(6/7) + 6/7 −3/7 = 0

𝑟 2, 𝑟 3 = −6/7 2/7 + (3/7)(6/7) + 2/7 −3/7 = 0

𝑟1 = 𝑟1 , 𝑟1 1/2 = 3/7 2 + 2/7 2 + 6/7 2 = 1

𝑟 2 = 𝑟 2, 𝑟 21/2 = −6/7 2 + 3/7 2 + 2/7 2 = 1

∴ r 1, r 2, r 3 is an orthonormal set. Hence, it is an orthogonal matrix

𝑟 3 = 𝑟 3, 𝑟 31/2 = 2/7 2 + 6/7 2 + −3/7 2 = 1

1(b)

𝐶𝑜𝑙𝑢𝑚𝑛 𝑠𝑝𝑎𝑐𝑒 𝑜𝑓 𝐴: 𝑐 1 =

3/7−6/72/7

,

𝑐 2 =

2/73/76/7

, 𝑐 3 =

6/72/7

−3/7

𝑐 1, 𝑐 2 = 3/7 2/7 + (−6/7)(3/7) + 2/7 6/7 = 0

𝑐 1, 𝑐 3 = 3/7 6/7 + (−6/7)(2/7) + 2/7 −3/7 = 0

𝑐 2, 𝑐 3 = 2/7 6/7 + (3/7)(2/7) + 6/7 −3/7 = 0

𝑐1 = 𝑟1 , 𝑟1 1/2 = 1

𝑐 2 = 𝑟 2, 𝑟 21/2 = 1

∴ c 1, c 2, c 3 is an orthonormal set. Hence, it is an orthogonal matrix

𝑐 3 = 𝑟 3, 𝑟 31/2 = 1

1(c)

2. Inverse of matrix A

𝐴 𝐼 𝐼 𝐴−1

3/7 2/7 6/7−6/7 3/7 2/72/7 6/7 −3/7

1 0 00 1 00 0 1

𝑖 7𝑅1 , 7𝑅2 , 7𝑅3

𝑖𝑖 𝑅1 − 𝑅2, 2𝑅1 + 𝑅2 , 𝑅2 + 3𝑅3

𝑖𝑖𝑖 𝑅2

7, 3𝑅2 − 𝑅3

𝑖𝑣 𝑅3

49

𝑣 𝑅2 − 2𝑅3

𝑣𝑖 𝑅1 + 4𝑅2 − 9𝑅3 1 0 00 1 00 0 1

3/7 −6/7 2/72/7 3/7 6/76/7 2/7 −3/7

∴ A is an orthogonal matrix from question 1 𝐴𝑇 = 𝐴−1

Transition matrix B to B′ 3(a)

𝑀𝐵 =1 23 2

, 𝑀𝐵′ =2 31 −4

𝑀𝐵′ 𝑀𝐵 𝐼 𝑃

2 31 −4

1 23 2

𝑖 𝑅1 ↔ 𝑅2 𝑖𝑖 2𝑅1 − 𝑅2

𝑖𝑖𝑖 −𝑅2

11 𝑖𝑣 𝑅1 + 4𝑅2

1 00 1

13/11 14/11−5/11 −2/11

∴ Transition matrix , P =13/11 14/11−5/11 −2/11

Transition matrix B′ to B 3(b)

1 23 2

2 31 −4

𝑖 𝑅2 − 3𝑅1

𝑖𝑖 −𝑅2

4

𝑖𝑖𝑖 𝑅1 − 2𝑅2

1 00 1

−1/2 −7/25/4 13/4

∴ Transition matrix , Q =−1/2 −7/25/4 13/4

𝑀𝐵 𝑀𝐵′ 𝐼 𝑄

Find 𝑤 𝐵′ 𝑖𝑓 𝑤 𝐵 =−13

3(c)

using equation 𝑣 𝐵′ = 𝑃 𝑣 𝐵

𝑤 𝐵′ = 𝑃 𝑤 𝐵

𝑤 𝐵′ =13/11 14/11−5/11 −2/11

−13

𝑀𝐵′ 𝑀𝐵 𝐼 𝑃

Transition matrix , P =13/11 14/11−5/11 −2/11

From 3(a)

𝑤 𝐵′ =29/11−1/11

Transition matrix B to B′ 4(a)

𝑀𝐵 =1 −1 −12 −1 11 1 7

, 𝑀𝐵′ =3 0 17 4 0

−2 1 0 𝑀𝐵′ 𝑀𝐵 𝐼 𝑃

3 0 17 4 0

−2 1 0 1 −1 −12 −1 11 1 7

1 0 00 1 00 0 1

−2/15 −1/3 −9/511/15 1/3 17/57/5 0 22/5

∴ Transition matrix , P

Transition matrix B′ to B 4(b)

1 −1 −12 −1 11 1 7

3 0 17 4 0

−2 1 0

∴ Transition matrix , Q =

11 11 −423/2 29/2 −13/2−7/2 −7/2 3/2

𝑀𝐵 𝑀𝐵′ 𝐼 𝑄

ERO

ERO 1 0 00 1 00 0 1

11 11 −423/2 29/2 −13/2−7/2 −7/2 3/2

Find w B′ first and then find w B 4(c)

using equation 𝑣 𝐵 = 𝑃−1 𝑣 𝐵′

𝑤 𝐵 = 𝑃−1 𝑤 𝐵′

𝑤 𝐵 =

11 11 −423/2 29/2 −13/2−7/2 −7/2 3/2

−22/1516/1512/5

𝑀𝐵′ 𝑀𝐵 𝐼 𝑃

Transition matrix , P =13/11 14/11−5/11 −2/11

From 4(a)

𝑤 𝐵 =−14−175

𝑀𝐵′ 𝑤 𝐼 w B′

3 0 17 4 0

−2 1 0 −2−64

1 0 00 1 00 0 1

−22/1516/1512/5

ERO

w B′

𝑤 𝐵 = 𝑄 𝑤 𝐵′

𝑀𝐵 𝑤 𝐼 w B

1 −1 −12 −1 11 1 7

−2−64

1 0 00 1 00 0 1

−14−175

−2𝑅1 + 𝑅2

Find w B directly 4(d)

∴ 𝑤 𝐵 =−14−175

1 −1 −10 1 30 2 8

−2−26

−𝑅1 + 𝑅3

𝑅1 + 𝑅2 1 0 20 1 30 0 2

−4−210

−2𝑅2 + 𝑅3

𝑅3/2 1 0 20 1 30 0 1

−4−25

𝑅1 − 2𝑅2

𝑅2 − 3𝑅3

Transition matrix B to B′ 5(a)

𝑀𝐵 =3 −41 2

, 𝑀𝐵′ =2 10 3

𝑀𝐵′ 𝑀𝐵 𝐼 𝑃

2 10 3

3 −41 2

𝑖 𝑅1/2

𝑖𝑖 𝑅2/3

𝑖𝑖𝑖 𝑅1 −𝑅2

2

1 00 1

4/3 −7/31/3 2/3

∴ Transition matrix , P =4/3 −7/31/3 2/3

Find q B first and then find q B′ 5(b)

using equation 𝑣 𝐵′ = 𝑃 𝑣 𝐵

𝑞 𝐵′ = 𝑃 𝑞 𝐵

𝑀𝐵 𝑞 𝐼 q B

3 −41 2

−14

1 00 1

7/5

13/10 ERO

q B

𝑞 𝐵′ =4/3 −7/31/3 2/3

7/513/10

𝑞 𝐵′ =−7/64/3

𝑀𝐵′ 𝑞 𝐼 q B′

2 10 3

−14

𝑅1/2

Find q B′ directly 5(c)

𝑅2/3 1 1/20 1

−1/24/3

−𝑅2

2+ 𝑅1

1 00 1

−7/64/3

∴ 𝑞 𝐵′ =−7/64/3